A reaction, $A + B \rightarrow C + D + q$ is found to have a positive entropy change. The reaction will be (i) possible at high temperature (ii) possible only at low temperature (iii) not possible at any temperature (v) possible at any temperature

A reaction, A + B arrow C + D + q is found to have a positive entropy change. The reaction will be (i) possibl

Exercises · 5.6

A + B \rightarrow C + D + q

(i) possible at high temperature

(ii) possible only at low temperature

(iii) not possible at any temperature

(v) possible at any temperature

Yanam CbseNCERTSubjective· 1mImportance★★★★★est

20% · 20/98 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Entropy Change Prediction

Entropy Change Prediction

Imagine you have a box of marbles — all neatly arranged, reds on one side, blues on the other. Now shake the box. What happens? The colours mix. They never spontaneously unmix. That tendency — for things to go from ordered to disordered — is what entropy measures. Entropy is a measure of disorder or randomness in a system.

When you predict an entropy change, you're asking: Will this process make the system more disordered or less disordered? And by how much?

The Core Intuition

Entropy change ( $\Delta S$ ) is positive when disorder increases, negative when disorder decreases. Three things drive this:

Volume change — More space means more positions for particles → more disorder. A gas expanding into vacuum has $\Delta S > 0$ .
Temperature change — Higher temperature means particles move faster, explore more states → more disorder. Heating something increases entropy.
Phase change — Solid → liquid → gas is a ladder of increasing disorder. Melting ice increases entropy; freezing water decreases it.

Note

Entropy always increases for spontaneous processes in an isolated system (Second Law of Thermodynamics). But for a non-isolated system, entropy can decrease locally — as long as the surroundings' entropy increases enough to compensate.

The Precise Statement

For a reversible process at constant temperature, the entropy change is:

\Delta S = \frac{Q_{\text{rev}}}{T}

where $Q_{\text{rev}}$ is the heat transferred reversibly, and $T$ is the absolute temperature (in Kelvin).

For an irreversible process (which is what actually happens), you calculate $\Delta S$ by imagining a reversible path between the same initial and final states — because entropy is a state function. It depends only on where you start and end, not how you get there.

\Delta S = \int \frac{dQ_{\text{rev}}}{T}

For common cases, you use these:

Process	Formula	Sign intuition
Isothermal expansion/compression (ideal gas)	$\Delta S = nR \ln\frac{V_2}{V_1}$	Expanding → more volume → $\Delta S > 0$
Heating/cooling (constant pressure)	$\Delta S = nC_p \ln\frac{T_2}{T_1}$	Heating → higher T → $\Delta S > 0$
Phase change (melting, boiling)	$\Delta S = \frac{\Delta H_{\text{phase}}}{T_{\text{phase}}}$	Melting/boiling → more disorder → $\Delta S > 0$

How to Predict Without Calculation

You don't always need numbers. Ask these questions in order:

Is there a phase change? Solid → liquid or liquid → gas always increases entropy. Reverse decreases it.
Is the number of gas molecules changing? In a reaction, more gas molecules means more disorder. $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ (gas → liquid) has $\Delta S < 0$ because 3 gas molecules become 2 liquid molecules.
Is temperature increasing or decreasing? Higher temperature → higher entropy.
Is volume increasing? More space → more entropy. …

Start your 14-day free trial to unlock the full solution →

A reaction with positive entropy change ( $\Delta S > 0$ ) and negative enthalpy change ( $\Delta H < 0$ , since $q$ is heat released) is spontaneous at all temperatures because $\Delta G = \Delta H - T\Delta S$ is always negative. The correct option is (v).

The key to this question lies in the Gibbs free energy criterion for spontaneity. For any process at constant temperature and pressure, the reaction is spontaneous when $\Delta G < 0$ , where $\Delta G = \Delta H - T\Delta S$ . The sign of $\Delta G$ depends on the signs and relative magnitudes of $\Delta H$ and $\Delta S$ , and on the temperature $T$ .

Here, the reaction $A + B \rightarrow C + D + q$ releases heat $q$ . That means the system loses energy to the surroundings — the reaction is exothermic. For an exothermic reaction, the enthalpy change $\Delta H$ is negative. The problem also states that the entropy change $\Delta S$ is positive. So we have:

$\Delta H < 0$ (negative)
$\Delta S > 0$ (positive)

Now, plug these into the Gibbs equation:

\Delta G = \Delta H - T\Delta S

Since $\Delta H$ is negative and $-T\Delta S$ is also negative (because $T > 0$ and $\Delta S > 0$ ), the sum of two negative terms is always negative, regardless of the temperature. Let’s check each temperature regime:

At low temperatures: The term $T\Delta S$ is small, so $\Delta G \approx \Delta H$ (negative). The reaction is spontaneous.
At high temperatures: The term $T\Delta S$ becomes large, but it is subtracted from a negative $\Delta H$ . So $\Delta G$ becomes even more negative. The reaction remains spontaneous. …

Start your 14-day free trial to unlock the full solution →

←Previous question · exercises5.5 — The enthalpy of combustion of methane, graphite and dihydrogen at 298 …→Next question · exercises5.7 — In a process, 701 J of heat is absorbed by a system and 394 J of work …

Q.A reaction, A+B→C+D+qA + B \rightarrow C + D + qA+B→C+D+q is found to have a positive entropy change. The reaction will be

🎓Unlock everything free for 14 days

Q.A reaction, $A + B \rightarrow C + D + q$ is found to have a positive entropy change. The reaction will be

Unlock everything free for 14 days